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lina2011 [118]
3 years ago
9

Which descriptor applies to the circuit? A- on B- closed C- incomplete D- open

Physics
2 answers:
gayaneshka [121]3 years ago
8 0
Answer is D- open.hope you get it
HACTEHA [7]3 years ago
3 0
Open circuit is the answer
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Which of Newton's laws explains why satellites need very little fuel to stay in oribit?
Angelina_Jolie [31]

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.


Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

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4 0
3 years ago
What did early experiments and Coulomb’s Law describe? Select all that apply.
Luda [366]
<span>Like charges repel and opposite charges attract.
The further away two charged objects are the weaker the electrical force between them.
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4 0
3 years ago
Read 2 more answers
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
Planets in our solar system do not revolve around the sun in perfect circles. Their orbits are more like ovals that scientists d
galben [10]
Planets in our solar system do not revolve around the sun in perfect circles. Their orbits are more like ovals that scientists describe as elliptical. It is one of Kepler's laws. The sun is the focus of all the planets. The correct answer is D.
7 0
4 years ago
Help please..................
Shtirlitz [24]
<h3><em>If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart. If two objects with opposite charges are brought towards each other the force will be attractive, it will pull them towards each other.</em></h3><h3><em>hope it helps.... thank you....</em></h3>
7 0
3 years ago
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